# Difference between revisions of "2011 AIME I Problems/Problem 6"

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− | If the vertex is at (1 | + | If the vertex is at <math>(\frac{1}{4}, -\frac{9}{8})</math>, the equation of the parabola can be expressed in the form <math>y=a(x-\frac{1}{4})^2-\frac{9}{8}</math>. |

− | Expanding, we find that y=a(x^2-x | + | Expanding, we find that <math>y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}</math> , and <math>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}</math>. From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</math>, where <math>a+b+c</math> is an integer. From the above equation, we can conclude that <math>a=a</math>, <math>-\frac{a}{2}=b</math>, and <math>\frac{a}{16}-\frac{9}{8}=c</math>. Adding up all of these gives us <math>\frac{9a-18}{16}=a+b+c</math>. We know that <math>a+b+c</math> is an integer, so 9a-18 must be divisible by 16. Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. Therefore, if <math>9a=2</math>, <math>a=\frac{2}{9}</math>. Adding up gives us <math>2+9=\boxed{011}</math> |

## Revision as of 18:39, 18 March 2011

If the vertex is at , the equation of the parabola can be expressed in the form . Expanding, we find that , and . From the problem, we know that the parabola can be expressed in the form , where is an integer. From the above equation, we can conclude that , , and . Adding up all of these gives us . We know that is an integer, so 9a-18 must be divisible by 16. Let . If , then . Therefore, if , . Adding up gives us